If x = `(4"t")/(1 + "t"^2)`, y = `3((1 - "t"^2)/(1 + "t"^2))`, then show that `("d"y)/("d"x) = (-9x)/(4y)`

#### Solution

x = `(4"t")/(1 + "t"^2)`

Differentiating both sides w.r.t. ‘t’, we get

`("d"x)/"dt" = "d"/"dt" ((4"t")/(1 + "t"^2))`

= `((1 + "t"^2)*"d"/"dt"(4"t") - 4"t"*"d"/"dt"(1 + "t"^2))/(1 + "t"^2)^2`

= `((1 + "t"^2)(4) - 4"t"(0 + 2"t"))/(1 + "t"^2)^2`

= `(4 + 4"t"^2 - 8"t"^2)/(1 + "t"^2)^2`

= `(4 - 4"t"^2)/(1 + "t"^2)^2`

= `(4(1 - "t"^2))/(1 + "t"^2)^2`

y = `3((1 - "t"^2)/(1 + "t"^2))`

`("d"y)/"dt" = 3*"d"/"dt"((1 - "t"^2)/(1 + "t"^2))`

= `3[((1 + "t"^2)*"d"/"dt"(1 - "t"^2) - (1 - "t"^2)*"d"/"dt"(1 + "t"^2))/(1 + "t"^2)]`

= `3[((1 + "t"^2)(0 - 2"t") - (1 - "t"^2)(0 + 2"t"))/(1 + "t"^2)^2]`

= `3[(-2"t"(1 + "t"^2) - 2"t"(1 - "t"^2))/(1 + "t"^2)^2]`

= `3(- 2"t")[(1 + "t"^2 + 1 - "t"^2)/(1 + "t"^2)^2]`

= `- 6"t" xx 2/(1 + "t"^2)^2`

= `(-12"t")/(1 + "t"^2)^2`

∴ `("d"y)/("d"x) = (("d"y)/("dt"))/(("d"x)/("dt"))`

= `((-12"t")/((1 + "t"^2)^2))/((4(1 - "t"^2))/((1 + "t"^2)^2)`

= `("d"y)/("d"x) = (-3"t")/(1 - "t"^2)` ......(i)

Also, `(-9x)/(4y) = (-9((4"t")/(1 + "t"^2)))/(4 xx 3((1 - "t"^2)/(1 + "t"^2))`

= `(-3"t")/(1 -"t"^2)` ......(ii)

From (i) and (ii), we get

`("d"y)/("d"x) = (-9x)/(4y)`